Answer
(a) d-glutamic acid
(b) l-leucine $0.576mW/m^2$; d-glutamic acid $0.732mW/m^2$
Work Step by Step
(a) We know that if the transmission axis of the first polarizer is vertical, then the second polarizer has a horizontal transmission axis and hence the light will be polarized vertically after the first polarizer. The intensity of the transmitted light depends on the amount of rotation towards the horizontal. As the d-glutamic acid takes more rotation as compered l-leucine, it will have the greater intensity.
(b) We know that
$I_t=\frac{I_{\circ}}{2}(90+0.55)$
$I_t=(0.576\times 10^{-3}W/m^2)(\frac{1000mW/m^2}{1W/m^2})$
$I_t=0.576mW/m^2$
and $I_d=\frac{I_{\circ}}{2}(90-0.620)$
$I_d=(0.732\times 10^{-3}W/m^2)(\frac{1000mW/m^2}{1W/m^2})$
$I_d=0.732mW/m^2$
